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### Topic: Problem of the week - 22/04/2013  (Read 15097 times)

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#### Borek

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##### Problem of the week - 22/04/2013
« on: April 22, 2013, 06:30:41 PM »
Not a difficult one, but definitely based on something else than all the previous ones.

Ultra pure water was put into a cell with electrodes both of exactly 2 cm2 surface and exactly 0.5 cm apart. At 283 K measured resistance was 8.772 MΩ. Estimate value of water ion product at this temperature, knowing that limiting molar conductivities are respectively

$$\Lambda^\infty_{H^+} = 275 \frac {cm^2} {\Omega ~mol}$$

and

$$\Lambda^\infty_{OH^-} = 140 \frac {cm^2} {\Omega ~mol}$$
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#### Sunil Simha

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##### Re: Problem of the week - 22/04/2013
« Reply #1 on: April 29, 2013, 01:13:48 AM »
Given that resistance is 8.772 MΩ,

R = l/(kA) where l is the length and A is the area of the cell. k is the conductivity.

I get k= 2.85 x 10-8 /Ωcm

So molar conductivity is Λ=k/c where c is the concentration. Here assuming 1g/cc as the density of water, I get c = (1/18) mol/cc.

This gives Λ = 51.3 x 10-8 cm2/Ωmol.

knowing that, I can find the degree of dissociation as α= Λ/Λ
Λ = 415 cm2/Ωmol.

So α = 1.23 x 10-9.

Thus the water product is (c*α)2 which is 4.67 x 10-15 mol2/L2.

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